Is the Real Part of (z+1)^100 and (z-1)^100 Always Zero for Complex z?

  • Thread starter Thread starter oab729
  • Start date Start date
  • Tags Tags
    Zero
oab729
Messages
12
Reaction score
0

Homework Statement


(z+1)^100=(z-1)^100 z is complex


Homework Equations





The Attempt at a Solution



(z-1)/(z+1)=e^(i2pi(k/100))
 
Physics news on Phys.org
Now use exp(ix)=cosx+isinx and multiply by the complex conjugate of the denominator.
 
Word. I think I was just tired. I feel dumb for not seeing that.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top